Question

Question: An electron passes location $\left(0.02,0.04,-0.06\right)\mathit{m}$, and $2\mu s$later is detected at location $\left(0.02,1.84,-0.86\right)\mathbf{}\mathit{m}$, (1 microsecond is$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{s}$). a) What is the average velocity of the electron? b) If the electron continues to travel at this average velocity, where will it be in another $5\mu s$?

Short answer

Answer

a) The average velocity of the ball is $\left(0,-900000,400000\right)m/s$

b) The electron will be at the position$\left(0.02,-0.86,0.34\right)m$.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The electron passes the location$\left(0.02,0.04,-0.06\right)\mathbf{}\mathit{m}$.

• The electron has been detected in the location$\left(0.02,1.84,-0.86\right)\mathbf{}\mathit{m}$.

The electron is detected after$2\mu s$.

Significance of Newton’s first law to calculate the average velocity and position

This law states that an object will continue to move in a uniform motion unless it is resisted by an external object.

Determination of the average velocity and distance

a) From Newton’s first law, the formula for the average velocity of the electron can be expressed as:

${\overrightarrow{v}}_{average}=\left(\frac{\left(x_B-x_A\right)}{\Delta t_{AB}},\frac{\left(y_B-y_A\right)}{\Delta t_{AB}},\frac{\left(z_B-z_A\right)}{\Delta t_{AB}}\right)$

Here, $t_{AB}$The time difference of a particle from location A to location B

$x_A,y_A,z_A=$The positions of the ball at A position in the x, y, and z coordinates

$x_B,y_{B,}{z}_B=$The positions of the ball at B position in the x, y, and z coordinate

Substituting the values localid="1662456388346" $t_{AB}=2\mu s,x_A,y_A,z_A=(0.02,0.04,-0.06)\text{m}$$and{x}_B,y_{B,}{z}_B=(0.02,1.84,-0.86)\text{m}$in the above equation, we get:

$$\begin{gathered} {\overrightarrow{v}}_{average}=\left(\frac{\left(0.02-0.02\right)\text{m}}{2\times {10}^{-6}\text{s}},\frac{\left(0.04-1.84\right)\text{m}}{2\times {10}^{-6}\text{s}},\frac{\left(-0.06+0.86\right)\text{m}}{2\times {10}^{-6}\text{s}}\right) \\ {\overrightarrow{v}}_{average}=\left(0,-900000,400000\right)\text{m/s} \end{gathered}$$

Thus, the average velocity of the ball is $\left(0,-900000,400000\right)\text{m/s}$.

b) From Newton’s first law, the equation of displacement of the ball in the x-coordinate can be expressed as:

$x_C=x_B+v_{average,x}·\Delta t_{BC}$

Here,$x_C=$The displacement of the particle in the C point in the x-quadrant,

$t_{BC}=$The displacement of the particle in the B point in the x-qua

$x_B=$The time difference of a particle from location B to location C

$v_{average,x}=$The average velocity of the particle in the localid="1662455496234" $x$quadrant

Substituting the value , localid="1662456406627" $t_{Bc}=3\mu s,v_{average,x}=0\text{m/s}and{x}_B=0.02\text{m}$n the above equation, we get:

$$\begin{gathered} x_C=0.02\text{m + 0}·\Delta {\text{t}}_{BC} \\ {x}_C=0.02\text{m} \end{gathered}$$

Similarly, the equation of displacement of the ball in the y coordinate can be expressed as:

$y_C=y_B+v_{average,y}·\Delta t_{BC}$

$y_C=$The displacement of the particle in the C point in the y -quadrant,

$y_B=$The displacement of the particle in the B point in the y -quadrant,

$t_{BC}=$The time difference of a particle from location B to location C

$v_{average,y}·$The average velocity of the particle in the y -quadrant

Substituting the value$t_{BC}=3\mu c,v_{average,y}=-900000\text{m/sand}{y}_B=1.84\text{m}$, in the above equation, we get:

$$\begin{gathered} {\text{y}}_{\text{C}}\text{=}\left(\text{1.84}\text{m}\right)\text{+}\left(\text{900000}\text{m/s}\right)\times \left(\text{3μ}\text{s}\right) \\ \text{= 1.84}\text{m - 2.7}\text{m} \\ \text{= - 0.86}\text{m} \end{gathered}$$

Similarly, the equation of displacement of the ball in the z coordinate can be expressed as:

$z_C=z_B+v_{average,z}·\Delta t_{BC}$

$z_C=$The displacement of the particle in the C point in the z-quadrant,

$z_B=$The displacement of the particle in the B point in the z -quadrant,

$t_{BC}=$The time difference of a particle from location B to location C

$v_{average,z}=$The average velocity of the particle in the z quadrant

Substituting the values $t_{BC}=3\mu s,v_{average,z}=400000\text{m/sand}{z}_B=-0.86\text{m}$in the above equation, we get:

$$\begin{gathered} z_C=\left(-0.86\text{m}\right)\text{+}\left(400000\text{m/s}\right)\times \left(\text{3s}\right) \\ =-0.86\text{m}+1.2\text{m} \\ =0.34\text{m} \end{gathered}$$

Thus, the electron will be at the position$(0.02,-0.86,0.34)\text{m}$.